A compound pipeline consists of two pieces of identical pipes. The equivalent length of the same diameter and the same friction factor for this compound pipeline is L₁ when the pipes are connected in series and is L₂ when the pipes are connected in parallel. Determine the ratio of equivalent lengths, that is, L₁ to L₂.
- A64 : 1
- B8 : 1Correct
- C4 : 1
- D32 : 1
Solution
Concept
In series the lengths add. In parallel the head loss is the same in each pipe and the flows add.
Formula
hf = f L Q2 / (12.1 D5)
Step-by-step solution
1. Let each pipe have length L, diameter D and friction factor f. 2. Series: the same Q flows through both pipes, so the equivalent length L₁ = L + L = 2L. 3. Parallel: each pipe carries Q/2, so hf ∝ L (Q/2)2 = L Q2 / 4. 4. An equivalent pipe carrying the full Q with the same hf needs L₂ Q2 = L Q2 / 4, so L₂ = L/4. 5. Ratio L₁ : L₂ = 2L : L/4 = 8 : 1.
Explanation
In series the two lengths simply add to give 2L. In parallel each pipe carries half the flow, so a single pipe with the full flow needs only L/4 to give the same head loss. The ratio is therefore 8 : 1.
Shortcut
For n identical pipes: series length = nL and parallel length = L/n2, so ratio = n3 = 8 for n = 2.
Correct answer: B) 8 : 1